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| Which Door do you choose? |
| Stick with door #1. |
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36% |
[ 15 ] |
| Change to door #2. |
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34% |
[ 14 ] |
| It doesn't matter, it's a 50/50 chance! |
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29% |
[ 12 ] |
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| Total Votes : 41 |
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atarom
Dalai Lama of RealPoor
Joined: 11 Oct 2002
Posts: 16398
Location: 375th st. Y
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 Posted: 10/29/05 - 10:39 Post subject: The Monty Hall Problem
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I've known the answer to this for a long while, and it made sense to me in the past, but it never made as much sense as it did after drawing a simple chart.
If you don't know it, it's based on the show "let's make a deal" and it follows:
You are given a choice of three doors. Behind 2 doors are goats. Behind 1 door is a new shiny car.
You choose door # 1, and without opening it, Monty Hall opens one door #3, revealing a goat. There are now 2 doors remaining and you are given the option to choose between the two. The question is this: Is it better to stick with your door #1, or to change to door #2.
I think I'll make this a poll, just for fun.
No cheating. Vote now.
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sinrakin
RealPoor Master of Posts
Joined: 11 Oct 2002
Posts: 7044
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Posted: 10/29/05 - 12:54 Post subject:
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It's a tricky paradox, but I think most of the difficulty is an extremely subtle ambiguity in how it's phrased, and how that relates to the definition of the problem space. The key point being what exactly is meant by "Monty Hall opens one door #3", or to phrase it better, "Monty Hall opens one door, say number three".
Of course the funny thing is that the majority of people seem to be picking the one answer of the three that's not plausible 
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FockTop
Sir Postalot
Joined: 23 Dec 2004
Posts: 1055
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Posted: 10/29/05 - 13:08 Post subject:
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Not sure that I get it but...
You are given the choice to choose one of the 3 door...its not stated anywere that you must open them.
So if I get it right your choice is made not by opening them but by making the choice of a door.
So when you at first choose door 1 the result is a goat at door 3 right?
meaning that only 2 doors remain, door number 2 and door number 3.
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FockTop
Sir Postalot
Joined: 23 Dec 2004
Posts: 1055
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Posted: 10/29/05 - 13:10 Post subject:
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So the result of choosing door 1 is a goat in door 3...it doesnt mean that you get a goat for door 3.
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sinrakin
RealPoor Master of Posts
Joined: 11 Oct 2002
Posts: 7044
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Posted: 10/29/05 - 13:20 Post subject:
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Yeah, you pick a door but don't open it.
To state it more precisely, you pick one of the three doors but don't open it, Monty must pick one of the other two doors (i.e. he can't choose not to pick a door), and he must open it to show you a goat (it's not guaranteed to be any specific door, but it is guaranteed to have a goat behind it).
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FockTop
Sir Postalot
Joined: 23 Dec 2004
Posts: 1055
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Posted: 10/29/05 - 13:28 Post subject:
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Yes but you wont see the result of your choice until you open it?
If so then its a 50% chance with the other 2 doors.
If so then it should be open a door not make the choice of a door.
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Occulis
RealPoor Jedi
Joined: 11 Oct 2002
Posts: 13293
Location: Moral Relativity Central
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Occulis
RealPoor Jedi
Joined: 11 Oct 2002
Posts: 13293
Location: Moral Relativity Central
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Occulis
RealPoor Jedi
Joined: 11 Oct 2002
Posts: 13293
Location: Moral Relativity Central
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kbarr
RealPoor Jedi
Joined: 05 Oct 2004
Posts: 11239
Location: New York, now go fuck off...
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Posted: 10/29/05 - 14:26 Post subject:
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The french piece of shit doesn't know who Monty Hall is, or how to speak english.
/spit
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Booker
RealPoor Guru
Joined: 12 Oct 2002
Posts: 2562
Location: Corvallis, Oregon
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Posted: 10/30/05 - 05:20 Post subject:
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im still baffled, what is the answer, and why.
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Devook
RealPoor Guru
Joined: 31 Mar 2004
Posts: 2374
Location: Ypsilanti or Troy, MI
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Posted: 10/30/05 - 14:43 Post subject:
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Look here.
The three possibilities for the Monty Hall Problem, each with an equal chance of happening:
(x) ( ) ( )
( ) (x) ( )
( ) ( ) (x)
Consider it. Say you picked door number 1 for all three of those situations. What happens when he opens an empty door for each situation? Say you picked door 2 for all three of those situations and he opens the empty door. What fraction of the time will you win if you stick with your original door? How often will you win if you switch after he opens an empty door for you?
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Cethan
Can't Stop Posting
Joined: 18 Oct 2002
Posts: 505
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Posted: 10/30/05 - 14:52 Post subject:
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Sticking with door # 1 gives you a 1/3 chance of getting the good prize.
Changing to door # 3 gives you a 1/2 chance of getting the good prize.
Bling.
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Devook
RealPoor Guru
Joined: 31 Mar 2004
Posts: 2374
Location: Ypsilanti or Troy, MI
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Posted: 10/30/05 - 14:56 Post subject:
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| Cethan wrote: | Sticking with door # 1 gives you a 1/3 chance of getting the good prize.
Changing to door # 3 gives you a 1/2 chance of getting the good prize.
Bling. |
Nope... This would be assuming that the empty door is opened only as a result of indiscriminately choosing from the two other doors and throwing out any results in which the door with the prize was accidentally opened.
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Silvermouse
RealPoor Jedi
Joined: 12 Oct 2002
Posts: 11015
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Posted: 10/30/05 - 17:37 Post subject:
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I don't get it. How does changing to another door increase your odds of winning? One door is the winner, the other two are not. Revealing one of the wrong answers shouldn't change the odds. It should still be 1-2.
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Xarpolis
RealPoor Guru
Joined: 15 Oct 2002
Posts: 2884
Location: Philly, PA
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Posted: 10/30/05 - 17:57 Post subject:
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Yeah... this one really doesn't make sense to me either. After knowing what is in the 3rd door, you have the option of 2 doors. 1 has the prize, the other does not. At that point, isn't it still a 50/50 chance you're going to pick the right door?
Maybe psychology plays into it as well, where more people are opt to choose the middle door feeling that one will have the prize, and to counter this they placed the prize in the other one.
Maybe in that case the results arn't 50/50 'cause more people tend to chose one door than another instictually, and as such people who follow statistics know the least common door of choice and place the prize in that one.
Does that make any sense?
Anyway, no matter how you look at it, there's no "right" answer, is there?
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Devook
RealPoor Guru
Joined: 31 Mar 2004
Posts: 2374
Location: Ypsilanti or Troy, MI
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Posted: 10/30/05 - 17:58 Post subject:
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| Silvermouse wrote: | | I don't get it. How does changing to another door increase your odds of winning? One door is the winner, the other two are not. Revealing one of the wrong answers shouldn't change the odds. It should still be 1-2. |
Look at the visual example I gave. Just by going through the problem in your head you should see how it works, but for the sake of doing it mathematically:
When you pick a door originally, there is a 2/3 chance that you choose the door incorrectly. The fact that one of the other doors is empty doesn't change this probability, you knew already that at least one of the other doors was empty. The chance that one of the doors that you didn't choose has the prize is still 2/3. Monty Hall shows you which of the other doors is empty. The probability that the door you picked has the prize is <i>still</i> 1/3. The probability that a door you didn't pick has the prize is <i>still</i> 2/3, however now you know that one of the doors that you didn't pick is empty. That means that the remaining unpicked door has a 2/3 chance of holding the prize while the door you originally picked only has a 1/3 chance.
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Devook
RealPoor Guru
Joined: 31 Mar 2004
Posts: 2374
Location: Ypsilanti or Troy, MI
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Posted: 10/30/05 - 17:59 Post subject:
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| Xarpolis wrote: | Yeah... this one really doesn't make sense to me either. After knowing what is in the 3rd door, you have the option of 2 doors. 1 has the prize, the other does not. At that point, isn't it still a 50/50 chance you're going to pick the right door?
Maybe psychology plays into it as well, where more people are opt to choose the middle door feeling that one will have the prize, and to counter this they placed the prize in the other one.
Maybe in that case the results arn't 50/50 'cause more people tend to chose one door than another instictually, and as such people who follow statistics know the least common door of choice and place the prize in that one.
Does that make any sense?
Anyway, no matter how you look at it, there's no "right" answer, is there? |
No. There definitely <i>is</i> a mathematically proven "right" answer.
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mixelplux
Luke Warm
Joined: 23 Nov 2002
Posts: 344
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Posted: 10/30/05 - 21:15 Post subject:
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Logical flaw, even though you "mathematical" prove it. The setup for your mathematical proof is flawed the same way.
Look at it this way, each door has a 1/3 chance of being the winner. Revealing one that isn't a winner doesn't change the original chance of either of the other doors being the winner. Simply impossible.
An example: door 1 - goat, door 2 - prize, door 3 - goat.
Scenario 1:
Pick door 1, you will always be shown door 3. You argue that now door 2 has 2/3 chance of being right.
Scenario 2:
Pick door 2, you might be shown either door 1 or 3. To keep it simple, you are shown door 3. You now argue that door 1 has a 2/3 chance of being right.
Does that make sense? No, I didn't think so.
The flaw in logic is that the chance of 1 door having the prize is dependant on another door. You being shown what is behind one door that is known to have a goat doesn't affect what is behind either of the other 2 doors.
Another example for this:
Lets say you don't pick a door at first, and one door is opened and reveals a goat. Now either door is a 50/50 chance of being the winner. Given that this is identical to the examples above, picking or not picking affects your chance how? It's the same.
The problem is that what you are shown is always incorrect. Your mathematical proof would only be valid if the door you are shown is purely random, and could show the prize - but never did for whatever freak statistical reasons.
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Devook
RealPoor Guru
Joined: 31 Mar 2004
Posts: 2374
Location: Ypsilanti or Troy, MI
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Posted: 10/30/05 - 23:42 Post subject:
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| mixelplux wrote: | Logical flaw, even though you "mathematical" prove it. The setup for your mathematical proof is flawed the same way.
Look at it this way, each door has a 1/3 chance of being the winner. Revealing one that isn't a winner doesn't change the original chance of either of the other doors being the winner. Simply impossible.
An example: door 1 - goat, door 2 - prize, door 3 - goat.
Scenario 1:
Pick door 1, you will always be shown door 3. You argue that now door 2 has 2/3 chance of being right.
Scenario 2:
Pick door 2, you might be shown either door 1 or 3. To keep it simple, you are shown door 3. You now argue that door 1 has a 2/3 chance of being right.
Does that make sense? No, I didn't think so.
The flaw in logic is that the chance of 1 door having the prize is dependant on another door. You being shown what is behind one door that is known to have a goat doesn't affect what is behind either of the other 2 doors.
Another example for this:
Lets say you don't pick a door at first, and one door is opened and reveals a goat. Now either door is a 50/50 chance of being the winner. Given that this is identical to the examples above, picking or not picking affects your chance how? It's the same.
The problem is that what you are shown is always incorrect. Your mathematical proof would only be valid if the door you are shown is purely random, and could show the prize - but never did for whatever freak statistical reasons. |
No... actually you are completely, completely wrong. In every sense. Really. I'm not sure why you would argue this problem when you obviously don't know what you are talking about. You're just going to have to trust me on this one.
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mixelplux
Luke Warm
Joined: 23 Nov 2002
Posts: 344
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Posted: 10/31/05 - 00:07 Post subject:
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| Devook wrote: | | mixelplux wrote: | Logical flaw, even though you "mathematical" prove it. The setup for your mathematical proof is flawed the same way.
Look at it this way, each door has a 1/3 chance of being the winner. Revealing one that isn't a winner doesn't change the original chance of either of the other doors being the winner. Simply impossible.
An example: door 1 - goat, door 2 - prize, door 3 - goat.
Scenario 1:
Pick door 1, you will always be shown door 3. You argue that now door 2 has 2/3 chance of being right.
Scenario 2:
Pick door 2, you might be shown either door 1 or 3. To keep it simple, you are shown door 3. You now argue that door 1 has a 2/3 chance of being right.
Does that make sense? No, I didn't think so.
The flaw in logic is that the chance of 1 door having the prize is dependant on another door. You being shown what is behind one door that is known to have a goat doesn't affect what is behind either of the other 2 doors.
Another example for this:
Lets say you don't pick a door at first, and one door is opened and reveals a goat. Now either door is a 50/50 chance of being the winner. Given that this is identical to the examples above, picking or not picking affects your chance how? It's the same.
The problem is that what you are shown is always incorrect. Your mathematical proof would only be valid if the door you are shown is purely random, and could show the prize - but never did for whatever freak statistical reasons. |
No... actually you are completely, completely wrong. In every sense. Really. I'm not sure why you would argue this problem when you obviously don't know what you are talking about. You're just going to have to trust me on this one. |
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Devook
RealPoor Guru
Joined: 31 Mar 2004
Posts: 2374
Location: Ypsilanti or Troy, MI
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Posted: 10/31/05 - 00:36 Post subject:
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| mixelplux wrote: | | Devook wrote: | | mixelplux wrote: | Logical flaw, even though you "mathematical" prove it. The setup for your mathematical proof is flawed the same way.
Look at it this way, each door has a 1/3 chance of being the winner. Revealing one that isn't a winner doesn't change the original chance of either of the other doors being the winner. Simply impossible.
An example: door 1 - goat, door 2 - prize, door 3 - goat.
Scenario 1:
Pick door 1, you will always be shown door 3. You argue that now door 2 has 2/3 chance of being right.
Scenario 2:
Pick door 2, you might be shown either door 1 or 3. To keep it simple, you are shown door 3. You now argue that door 1 has a 2/3 chance of being right.
Does that make sense? No, I didn't think so.
The flaw in logic is that the chance of 1 door having the prize is dependant on another door. You being shown what is behind one door that is known to have a goat doesn't affect what is behind either of the other 2 doors.
Another example for this:
Lets say you don't pick a door at first, and one door is opened and reveals a goat. Now either door is a 50/50 chance of being the winner. Given that this is identical to the examples above, picking or not picking affects your chance how? It's the same.
The problem is that what you are shown is always incorrect. Your mathematical proof would only be valid if the door you are shown is purely random, and could show the prize - but never did for whatever freak statistical reasons. |
No... actually you are completely, completely wrong. In every sense. Really. I'm not sure why you would argue this problem when you obviously don't know what you are talking about. You're just going to have to trust me on this one. |
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Seriously. Google it or read that thing Occulis posted. I'm sure you'll find out that I'm right. I love when people try to debate the answer to a math problem, like when Brash tried to argue that gambling thing about infinitesimal probability. Mathematics is a science with definite answers, even when dealing with probabilities. There's no debate.
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kireol
RealPoor Master of Posts
Joined: 02 Aug 2003
Posts: 9517
Location: Royal Oak, MI
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Posted: 10/31/05 - 00:49 Post subject:
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unless I'm reading this incorrectly, it's 50/50. there's only 2 left. the 3rd is taken out of the equation.
It's like saying....you flip a quarter 24 times and it comes up heads every time. what are the odds of it being heads again? It's still 50/50.
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Silvermouse
RealPoor Jedi
Joined: 12 Oct 2002
Posts: 11015
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Posted: 10/31/05 - 01:58 Post subject:
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| Devook wrote: | | Silvermouse wrote: | | I don't get it. How does changing to another door increase your odds of winning? One door is the winner, the other two are not. Revealing one of the wrong answers shouldn't change the odds. It should still be 1-2. |
Look at the visual example I gave. Just by going through the problem in your head you should see how it works, but for the sake of doing it mathematically:
When you pick a door originally, there is a 2/3 chance that you choose the door incorrectly. The fact that one of the other doors is empty doesn't change this probability, you knew already that at least one of the other doors was empty. The chance that one of the doors that you didn't choose has the prize is still 2/3. Monty Hall shows you which of the other doors is empty. The probability that the door you picked has the prize is <i>still</i> 1/3. The probability that a door you didn't pick has the prize is <i>still</i> 2/3, however now you know that one of the doors that you didn't pick is empty. That means that the remaining unpicked door has a 2/3 chance of holding the prize while the door you originally picked only has a 1/3 chance. |
Why does it remain 1/3 if there are only two doors left?
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Devook
RealPoor Guru
Joined: 31 Mar 2004
Posts: 2374
Location: Ypsilanti or Troy, MI
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Posted: 10/31/05 - 02:19 Post subject:
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| Silvermouse wrote: | | Devook wrote: | | Silvermouse wrote: | | I don't get it. How does changing to another door increase your odds of winning? One door is the winner, the other two are not. Revealing one of the wrong answers shouldn't change the odds. It should still be 1-2. |
Look at the visual example I gave. Just by going through the problem in your head you should see how it works, but for the sake of doing it mathematically:
When you pick a door originally, there is a 2/3 chance that you choose the door incorrectly. The fact that one of the other doors is empty doesn't change this probability, you knew already that at least one of the other doors was empty. The chance that one of the doors that you didn't choose has the prize is still 2/3. Monty Hall shows you which of the other doors is empty. The probability that the door you picked has the prize is <i>still</i> 1/3. The probability that a door you didn't pick has the prize is <i>still</i> 2/3, however now you know that one of the doors that you didn't pick is empty. That means that the remaining unpicked door has a 2/3 chance of holding the prize while the door you originally picked only has a 1/3 chance. |
Why does it remain 1/3 if there are only two doors left? |
Because the probability of the prize being behind a particular door is 1/3. It doesn't matter whether you know that one of the other doors is open. One in every three doors has a prize behind it. Ignore the fact that there are only two doors left. The main point is that you are picking between your original choice or all the other choices.
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Silvermouse
RealPoor Jedi
Joined: 12 Oct 2002
Posts: 11015
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Posted: 10/31/05 - 02:43 Post subject:
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| Devook wrote: | | Silvermouse wrote: |
Why does it remain 1/3 if there are only two doors left? |
Because the probability of the prize being behind a particular door is 1/3. It doesn't matter whether you know that one of the other doors is open. One in every three doors has a prize behind it. Ignore the fact that there are only two doors left. The main point is that you are picking between your original choice or all the other choices. |
Well, I obviously won't pick the s****y door, which is door 3. So that leaves door 1 and door 2. One is a winner, one is not. I don't see why switching to door 2 would give me a better chance than the door I already picked.
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Silvermouse
RealPoor Jedi
Joined: 12 Oct 2002
Posts: 11015
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Posted: 10/31/05 - 03:12 Post subject:
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Here are two quotes from the original link that show both sides. First, is the side that represented my view:
| Quote: | Given 2 doors, one with a prize and one without, which should you pick?
I think you'd agree that odds are 50-50 you'll be right. The fact that I may have had 3 doors initially and opened one, or 1000 doors and opened 998, does not change the current problem. There is still only a 50/50 chance of winning.
Because the probability of the problem can be expressed in three different ways.
P1 - 1/3 chance prize is door #1, 2/3 chance it is not
P2 - 1/3 chance prize is door #2, 2/3 chance it is not
P3 - 1/3 chance prize is door #3, 2/3 chance it is not
Most of the logic I've seen has the following flaw. As soon as the contestant picks door #1, they immediately assume that P2 and P3 are not valid. Well, they are. So they follow only P1. Regardless of which door I pick, all 3 are still valid. It is only when Monty opens a door that things change. By opening door #3 we now have
P3 - 0/0 chance prize is door #3, 3/3 chance it is not.
Now here everyone says, "Hey by P1, then the 2/3 chance all goes to door #2' But then why not, "By P2, then the 2/3 chance all goes to door #1"? Both are incorrect. When Monty picks #3, the whole problem is changed to:
P1 - 1/2 chance prize is door #1, 1/2 chance it is not
P2 - 1/2 chance prize is door #2, 1/2 chance it is not
P3 - 0 chance prize is door #3, 1 chance it is not
Look at it another way.
What's being said is that if I pick #1 and monty shows #3, then 2/3 of the time #2 will win. So If I had picked #2 to start with and monty opens #3, then the odds go 2/3 to #1. Why would they change? |
Here is the other side of the problem, which makes sense as well!
| Quote: | | When n is 3 as in the original problem, things get confusing but now let n be a large number say 1000. So there are 999 empty doors.I select one of them and the host shows me 998 empty doors. Now it is clear that it is definitely advantageous for me to switch. There was only a 0.001 chance of me picking the correct door immediately so obviously one can't say that it is now a 50-50 chance. It is still 0.001 that I had picked the right door initially. So the chance of me winning if I now switch is 0.999 or (n-1)/n. Since this is the solution for the general problem of n doors, the answer for n=3 is (3-1)/2 or there is a 2/3 chance of winning if I switch my choice of doors. |
I think I understand now. You have a 1/3 of a chance from the start, and that chance never changes. If there were 12 doors and you picked 1, you'd have 1/12 of a chance to win if you stay, even if he reveals 10 loser doors and leaves you with 2. If there were a billion doors and you pick 1, you have a .000000001 chance of being right. Monty reveals 999,999,998 leaving you with two doors, you should switch because if you stay it means you're stuck with your original .000000001 chance, but each door he eliminates increases the odds that the door you didn't pick was the right one.
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Devook
RealPoor Guru
Joined: 31 Mar 2004
Posts: 2374
Location: Ypsilanti or Troy, MI
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Posted: 10/31/05 - 03:56 Post subject:
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| Silvermouse wrote: | Given 2 doors, one with a prize and one without, which should you pick?
I think you'd agree that odds are 50-50 you'll be right. The fact that I may have had 3 doors initially and opened one, or 1000 doors and opened 998, does not change the current problem. There is still only a 50/50 chance of winning.
Because the probability of the problem can be expressed in three different ways.
P1 - 1/3 chance prize is door #1, 2/3 chance it is not
P2 - 1/3 chance prize is door #2, 2/3 chance it is not
P3 - 1/3 chance prize is door #3, 2/3 chance it is not
Most of the logic I've seen has the following flaw. As soon as the contestant picks door #1, they immediately assume that P2 and P3 are not valid. Well, they are. So they follow only P1. Regardless of which door I pick, all 3 are still valid. It is only when Monty opens a door that things change. By opening door #3 we now have
P3 - 0/0 chance prize is door #3, 3/3 chance it is not.
Now here everyone says, "Hey by P1, then the 2/3 chance all goes to door #2' But then why not, "By P2, then the 2/3 chance all goes to door #1"? Both are incorrect. When Monty picks #3, the whole problem is changed to:
P1 - 1/2 chance prize is door #1, 1/2 chance it is not
P2 - 1/2 chance prize is door #2, 1/2 chance it is not
P3 - 0 chance prize is door #3, 1 chance it is not
Look at it another way.
What's being said is that if I pick #1 and monty shows #3, then 2/3 of the time #2 will win. So If I had picked #2 to start with and monty opens #3, then the odds go 2/3 to #1. Why would they change? |
The reason this logic is bad is because it assumes that once the empty door is opened, it's inconsequential, but the fact that Monty Hall opened <i>that</i> particular door means that the other door that he could have opened but didn't has a higher chance of not being empty than the door you picked, which he can't consider.
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Silvermouse
RealPoor Jedi
Joined: 12 Oct 2002
Posts: 11015
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Posted: 10/31/05 - 03:59 Post subject:
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| Devook wrote: | | Silvermouse wrote: | Given 2 doors, one with a prize and one without, which should you pick?
I think you'd agree that odds are 50-50 you'll be right. The fact that I may have had 3 doors initially and opened one, or 1000 doors and opened 998, does not change the current problem. There is still only a 50/50 chance of winning.
Because the probability of the problem can be expressed in three different ways.
P1 - 1/3 chance prize is door #1, 2/3 chance it is not
P2 - 1/3 chance prize is door #2, 2/3 chance it is not
P3 - 1/3 chance prize is door #3, 2/3 chance it is not
Most of the logic I've seen has the following flaw. As soon as the contestant picks door #1, they immediately assume that P2 and P3 are not valid. Well, they are. So they follow only P1. Regardless of which door I pick, all 3 are still valid. It is only when Monty opens a door that things change. By opening door #3 we now have
P3 - 0/0 chance prize is door #3, 3/3 chance it is not.
Now here everyone says, "Hey by P1, then the 2/3 chance all goes to door #2' But then why not, "By P2, then the 2/3 chance all goes to door #1"? Both are incorrect. When Monty picks #3, the whole problem is changed to:
P1 - 1/2 chance prize is door #1, 1/2 chance it is not
P2 - 1/2 chance prize is door #2, 1/2 chance it is not
P3 - 0 chance prize is door #3, 1 chance it is not
Look at it another way.
What's being said is that if I pick #1 and monty shows #3, then 2/3 of the time #2 will win. So If I had picked #2 to start with and monty opens #3, then the odds go 2/3 to #1. Why would they change? |
The reason this logic is bad is because it assumes that once the empty door is opened, it's inconsequential, but the fact that Monty Hall opened <i>that</i> particular door means that the other door that he could have opened but didn't has a higher chance of not being empty than the door you picked, which he can't consider. |
Yes, I figured that out in my post right above yours. It all makes sense now! Everybody switch!
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atarom
Dalai Lama of RealPoor
Joined: 11 Oct 2002
Posts: 16398
Location: 375th st. Y
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Posted: 10/31/05 - 09:29 Post subject:
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hehe this is why i posted it. it's counter intuitive. look. i woll draw you an easily understandable flow chart.
now you can all understand!
See! two times out of three, you switch, you get a car.
no wonder the show went off the air.
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